To correct for background, measure the count rate with no source present, then subtract that rate from every reading taken with the source. The result is the count rate that comes from the source alone. It appears whenever a half-life question gives you raw detector readings.
This skill opens the topic half-life and background. It follows the ideas in interpreting random count-rate data, where you saw that counts fluctuate.
Why does a detector record counts with no source?
Radiation from rocks, soil, building materials, the air and cosmic rays passes through the room all the time. A Geiger-Müller tube or similar detector counts some of it. Your source adds extra counts on top.
So a measured count rate has two parts: measured rate = source rate + background rate. The source rate is what the question really wants, so you rearrange to source rate = measured rate − background rate.
How do I correct a reading, step by step?
- Find the background rate. Divide the background counts by the time they were counted for, for example counts ÷ minutes.
- Check the units of the source readings. Are they already counts per minute, or are they a total over some time?
- Match the time. If the source reading is a total, multiply the background rate by the same counting time to get the background total for that time.
- Subtract. Corrected = measured − background, in the same units.
- Label the answer with the unit, such as counts/min.
Worked example
(Invented data.) With the source removed, a detector recorded 60 counts in 15 minutes. A source was then placed near it and the count rate was measured at three times.
| Time (min) | Measured count rate (counts/min) |
|---|---|
| 0 | 84 |
| 10 | 44 |
| 20 | 24 |
Step 1, background rate: 60 ÷ 15 = 4 counts/min.
Step 2, subtract from each reading:
| Time (min) | Corrected count rate (counts/min) |
|---|---|
| 0 | 84 − 4 = 80 |
| 10 | 44 − 4 = 40 |
| 20 | 24 − 4 = 20 |
Check: the corrected rate halves every 10 minutes (80, 40, 20), which is the pattern a half-life produces. The raw values do not: 84 to 44 is not half. Correction reveals the pattern. You will use it in determining half-life from a decay graph.
The mistake to watch for
The same source was also counted for 5 minutes and gave a total of 400 counts. A student wrote:
Mistaken answer: 400 − 60 = 340 counts
The student subtracted the background total for 15 minutes from a source total for 5 minutes.
The two totals cover different times. The background in 5 minutes is 4 × 5 = 20 counts.
The correct answer is 400 − 20 = 380 counts, which is 380 ÷ 5 = 76 counts/min. Always ask “what time does each number cover?” before you subtract.
Check yourself
1. A detector records 45 counts in 15 minutes with no source. With a source, it records 250 counts in 10 minutes. What is the corrected count rate from the source?
Show answer
Background rate = 45 ÷ 15 = 3 counts/min. In 10 minutes the background gives 3 × 10 = 30 counts. Corrected total = 250 − 30 = 220 counts. Corrected rate = 220 ÷ 10 = 22 counts/min.
2. The background is 10 counts/min. Measured rates are 130, 70 and 40 counts/min at equal time steps. Find the corrected rates and say whether they halve each step.
Show answer
Corrected: 130 − 10 = 120, 70 − 10 = 60, 40 − 10 = 30. Each value is half of the one before, so yes, they halve at each step.
3. A student measures 5 counts/min from a source but the background is 6 counts/min. What does the subtraction suggest?
Show answer
5 − 6 = −1 counts/min, which is not a real activity. The source reading is not distinguishable from background, and random variation made it slightly low.
Where this leads next
With clean data you can read a half-life from a graph in determine half-life from a decay graph. The bounds and rounding explainer can help you decide how many figures a corrected rate deserves.
Plenty of students understand subtraction but lose marks on the timing step. Our teachers look for that pattern in online one-to-one Physics tuition.